What is the integral of sqrt(x^2-2x+2) ?

The integral of sqrt(x^2-2x+2) is 1/2 (x-1)sqrt(x^2-2x+2)+1/2 ln|x-1+sqrt(x^2-2x+2)|+C

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integral of sqrt(x^2-2x+2)

\int x^{2} - 2 x + 2 d x

\frac{1}{2} (x - 1) x^{2} - 2 x + 2 + \frac{1}{2} ln x - 1 + x^{2} - 2 x + 2 + C

Solution steps

\int x^{2} - 2 x + 2 d x

Complete the square

x^{2} - 2 x + 2 : (x - 1)^{2} + 1

= \int (x - 1)^{2} + 1 d x

Apply u-substitution

: \int u^{2} + 1 d u

= \int u^{2} + 1 d u

Apply Trigonometric Substitution

: \int sec^{3} (v) d v

= \int sec^{3} (v) d v

Apply Integral Reduction

= \frac{sec ^{2} ( v ) sin ( v )}{2} + \frac{1}{2} \cdot \int sec (v) d v

\int sec (v) d v = ln ∣ tan (v) + sec (v) ∣

= \frac{sec ^{2} ( v ) sin ( v )}{2} + \frac{1}{2} ln ∣ tan (v) + sec (v) ∣

Substitute back

= \frac{sec ^{2} ( arctan ( x - 1 ) ) sin ( arctan ( x - 1 ) )}{2} + \frac{1}{2} ln ∣ tan (arctan (x - 1)) + sec (arctan (x - 1)) ∣

Simplify

\frac{sec ^{2} ( arctan ( x - 1 ) ) sin ( arctan ( x - 1 ) )}{2} + \frac{1}{2} ln ∣ tan (arctan (x - 1)) + sec (arctan (x - 1)) ∣ : \frac{1}{2} (x - 1) x^{2} - 2 x + 2 + \frac{1}{2} ln x - 1 + x^{2} - 2 x + 2

= \frac{1}{2} (x - 1) x^{2} - 2 x + 2 + \frac{1}{2} ln x - 1 + x^{2} - 2 x + 2

Add a constant to the solution

= \frac{1}{2} (x - 1) x^{2} - 2 x + 2 + \frac{1}{2} ln x - 1 + x^{2} - 2 x + 2 + C

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What is the integral of sqrt(x^2-2x+2) ?
The integral of sqrt(x^2-2x+2) is 1/2 (x-1)sqrt(x^2-2x+2)+1/2 ln|x-1+sqrt(x^2-2x+2)|+C